I am going through the book Probability and Random Processes for Computer and Electrical Engineers for my own understanding. Example 4.23 states that,
Let $Z := X + Y$ where $X$ and $Y$ are independent random variables. The characteristic function of $Z$ is the product of the characteristic functions of $X$ and $Y$,
$$ \phi_Z(v) := \mathbb{E}[e^{jvZ}] = \mathbb{E}[e^{jv(X + Y)}] = \mathbb{E}[e^{jvX}e^{jvY}] = \mathbb{E}[e^{jvX}]\mathbb{E}[e^{jvY}] = \phi_X(v)\phi_Y(v) $$
Which is pretty easy to see because they are independent. I cannot derive the next box though, even though it seems simple. 4.16 says,
An immediate consequence of the previous example is that if $X$ and $Y$ are independent continuous random variables, then the density of their sum $Z = X + Y$ is the convolution of their densities (which is the inverse Fourier transform of the above example). $$f_Z(z) = \int_{-\infty}^{\infty}f_X(z - y)f(y) dy$$
Trying to do this myself, I get stuck in the inverse Fourier transform,
$$ \begin{aligned} \phi_Z(v) &= \phi_X(v)\phi_Y(v) \\ \end{aligned} $$
So this means that...
$$ \begin{aligned} f_Z(z) &= \phi_Z^{-1}(v) \\ &= \phi_X^{-1}(v)\phi_Y^{-1}(v) \\ &= \int e^{-jvz}\phi_X(v) dv \int e^{-jvz}\phi_Y(v) dv \\ &= \iint e^{-jvz}e^{jvx}f(x) dxdv \iint e^{-jvz}e^{jvy}f(y) dydv \\ &= \iint e^{-jv(x-z)}f(x) dxdv \iint e^{-jv(y-z)}f(y) dydv \\ &= \iint e^{-jv(z -y - z)}f(z - y) dydv \iint e^{-jv(y-z)}f(y) dydv \quad \text{change of vars.}\;\; x = z-y\\ &= \iint e^{jvy}f(z - y) dydv \iint e^{-jv(y-z)}f(y) dydv \\ &= \iint e^{jvy}e^{-jv(y-z)}f(z - y)f(y) dydv \\ &= \iint e^{jv(z)}f(z - y)f(y) dydv \\ \end{aligned} $$
I can see I have all the pieces there, but I am unsure where I went wrong or how to finish it so I get rid of the double integral and achieve the expression in the blockquotes. Does anyone know what to do?
Let $F_X$, $F_Y$, $F_Z$ be the CDF's of $X$, $Y$, $Z$ respectively. Then, for any $u\in \mathbb{R}$, \begin{align*} F_{Z} (u) &= \mathbb{P}(Z \leq u) = \mathbb{P}(X+Y \leq u) = \iint_{x+y \leq u} f_{X}(x) f_{Y}(y)dxdy \\ &= \int_{-\infty}^{\infty} f_{X}(x)\left(\int_{-\infty}^{u-x} f_{Y}(y)dy \right) dx = \int_{-\infty}^{\infty} f_{X}(x) F_Y(u-x) dx. \end{align*} So \begin{align*} f_Z(u) &= \frac{d}{du} \int_{-\infty}^{\infty} f_{X}(x) F_Y(u-x) dx = \int_{-\infty}^{\infty} f_{X}(x) \frac{d}{du} F_Y(u-x) dx \\ & = \int_{-\infty}^{\infty} f_{X}(x) f_Y(u-x) dx = \int_{-\infty}^{\infty} f_{X}(u-v) f_Y(v) dv. \end{align*}